# If the integral is zero, when is the integrand also zero?

If

$$\int_{-\infty}^{\infty} f(k) e^{i\mathbf{k}.\mathbf{r}} d\mathbf{k} =0$$

then is

$$\ f(k)=0$$ ?

Is it correct to say that this is an expansion in an orthonormal basis, $$\ e^{ik.r}$$ , and so linear independence demands that f(k) be zero for all k?

Mute
Homework Helper
Yep, that works. You can show it explicitly by multiplying both sides by $\exp(-i\mathbf{q} \cdot \mathbf{r})$ and integrating over r. You should find f(q) = 0. (Assuming f(q) is reasonably 'nice'. I suppose there could be pathological examples where f(q) is non-zero).

CompuChip
Homework Helper
I believe that the basis being complete without redundancy is also important. In other words: you can say f(k) = 0, if any function (in your function space) can be expanded in the basis, and in a unique way.

[Edit: probably you can show this to be the case for an orthonormal basis, because it is impossible to express any basis element as (infinite) sum of the others].

Stephen Tashi
If

$$\int_{-\infty}^{\infty} f(k) e^{i\mathbf{k}.\mathbf{r}} d\mathbf{k} =0$$

then is

$$\ f(k)=0$$ ?

Is it correct to say that this is an expansion in an orthonormal basis, $$\ e^{ik.r}$$ , and so linear independence demands that f(k) be zero for all k?

No. This is an integral transform, not an expansion of a function in some orthonormal basis. If f(k) was non-zero at only a finite number of points, you would still get 0 for the integral. You might be able to say f(k) = 0 "almost everywhere" or "except on a set of measure zero".